parametric integral relating to hyperbolic function

Suppose that $a$ is real number such that $0<a<1$, how can we calculate $$ I(a)=\int_0^\infty \big(1-\frac{\tanh ax}{\tanh x}\big)dx .$$ As for some speical cases, I can work out $I(1/2)=1$. Any suggestion to the integral above? Tks.

Expanding @Lucian's comment, if $a=\frac{p}{q}\in\mathbb{Q}$, we have: $$\begin{eqnarray*}I(a)&=&q\int_{0}^{+\infty}\left(1-\frac{\tanh px}{\tanh qx}\right)\,dx=q\int_{1}^{+\infty}\left(1-\frac{x^{2p}-1}{x^{2p}+1}\cdot\frac{x^{2q}+1}{x^{2q}-1}\right)\frac{dx}{x}\\&=&2q\int_{1}^{+\infty}\frac{x^{2q}-x^{2p}}{(x^{2p}+1)(x^{2q}-1)}\cdot\frac{dx}{x}=q\int_{0}^{+\infty}\frac{x^{2p}-x^{2q}}{(1+x^{2q})(1-x^{2p})}\cdot\frac{dx}{x}\\&=&\pi i q\sum_{z\in Z\cap \{\Im z>0\}}\operatorname{Res}\left(\frac{x^{2p}-x^{2q}}{x(1+x^{2q})(1-x^{2p})},x=z\right)\end{eqnarray*}$$ where $$Z=Z_p\cup Z_q=\left\{e^{\frac{\pi i}{p}},e^{\frac{2\pi i}{p}},\ldots,e^{\frac{(p-1)\pi i}{p}}\right\}\cup\left\{e^{\frac{\pi i}{4q}},e^{\frac{3\pi i}{4q}},\ldots,e^{\frac{(2q-1)\pi i}{4q}}\right\}.$$ However, I strongly doubt that a nice closed formula exists, since, for instance: $$I(3/5)=\frac{\pi}{15} \sqrt{\frac{154}{3}+\frac{18}{\sqrt{5}}-4 \sqrt{30 \left(5+\sqrt{5}\right)}}.$$

If you are interested in particular cases:

$$\begin{align} I\left( \frac15 \right) & = \sqrt{2-\frac{2}{\sqrt{5}}} \, \, \pi \\ I\left( \frac14 \right) & = \frac{\pi}{2} + 1 \\ I\left( \frac13 \right) & = \frac{\sqrt{3}}{3} \, \pi \\ I\left( \frac25 \right) & = \left(\sqrt{2+\frac{2}{\sqrt{5}}} - \frac54 \right) \pi \\ I\left( \frac12 \right) & = 1 \\ I\left( \frac35 \right) & = \frac{10\pi}{3\sqrt3} + \sqrt{1+\frac{2}{\sqrt5}} \, \pi- \sqrt{5+2\sqrt5} \, \pi \\ I\left( \frac23 \right) & = \frac{3\pi}{4} - \frac{\sqrt{3} \, \pi}{3} \\ I\left( \frac34 \right) & = \frac{8\sqrt{3} \,\pi}{27} - \frac{\pi}{2} + \frac13 \\ \end{align}$$

There are also some closed-form for $a \geq 1$:

$$\begin{align} I(1) & = 0 \\ I(2) & = -\frac{\pi}{4} \\ I(3) & = -\frac{2\sqrt{3} \, \pi}{9} \\ I(4) & = -\frac{\sqrt{2} \, \pi}{4} - \frac{\pi}{8} \\ I(5) & = -\frac{2}{5}\sqrt{1+ \frac{2}{\sqrt5}} \, \pi \\ I(6) & = -\frac{5 \pi}{12} - \frac{\sqrt{3} \, \pi}{9} \end{align}$$

I've got them by using CAS and some manual simplification.

@spanferkel

$$\begin{align} I\left(\frac 45 \right) & = -\frac{1}{40} \left(25+48 \sqrt{5-2 \sqrt{5}}-2 \sqrt{2050-352 \sqrt{5}-400 \sqrt{25-11 \sqrt{5}}}\right) \pi \\ I\left(\frac 16 \right) & = \frac{5 \pi }{3 \sqrt{3}}+1 \end{align}$$

The expressions are simplifications of sums with $\arctan$ terms.

@spanferkel

$$\begin{align} I\left( \frac 18 \right) & = 1+\sqrt{2}\,\pi \\ I\left( \frac 28 \right) & = \frac{\pi}{2} +1\\ I\left( \frac 38 \right) & = -{\frac {16}{27}}\,\sqrt {3}\,\pi+\sqrt {2}\,\pi+\frac 13 \\ I\left( \frac 48 \right) & = 1\\ I\left( \frac 58 \right) & = -\sqrt {2}\,\pi +{\frac {32}{125}}\,\sqrt {25+10\,\sqrt {5}}\,\pi -{\frac {16}{125}}\,\sqrt {25-10\,\sqrt {5}}\,\pi+\frac 15 \\ I\left( \frac 68 \right) & = {\frac {8}{27}}\,\sqrt {3}\,\pi -\frac{\pi}{2} + \frac 13 \end{align}$$

I couldn't find a closed-form of $I(7/8)$. The expressions are coming again from sums in term of $\arctan$.

There is a possible close-form in the website \${http://integralsandseries.prophpbb.com/topic482.html}$.

His result shows that $$ -\sum_{k=0}^{\infty}\Big(_k^{-2}\Big)\frac{\psi\big(a(k+1)\big)}{k+1}-\frac{2\ln 2}{a}-2\gamma.$$ Right? we can check it whatever.